Lecture 17 Solutions: Binary Stars & Stellar Masses
Practice Problem Solutions
Student note: These are model solutions written to show setup, units, and reasoning clearly. Your own work can be shorter as long as the physics, units, and logic are sound.
Use these after you try the problems yourself. The goal is to compare your reasoning to a clean worked example, not to skip the thinking step.
Core Problems
Problem 1: Combined Mass of a Visual Binary
Given: - \(P = 20\ \text{yr}\) - \(a = 4\ \text{AU}\)
Find: \(M_1 + M_2\) in solar masses
Equation: \[ M_1 + M_2 = \frac{a^3}{P^2} \]
This solar-unit form works when \(a\) is in AU, \(P\) is in years, and mass is in \(M_\odot\).
Solution:
\[ M_1 + M_2 = \frac{4^3}{20^2} = \frac{64}{400} = 0.16 \]
So
\[ M_1 + M_2 = 0.16\,M_\odot \]
Unit check: Because we used AU and years in the solar-unit form of Kepler’s law, the answer comes out directly in solar masses.
Sanity check: A wide binary with a fairly long period can easily have a low combined mass, so a value well below \(1\,M_\odot\) is plausible.
Answer: The combined mass is \(0.16\,M_\odot\).
Problem 2: Spectroscopic Mass Ratio
Restatement: We know the radial velocity amplitudes of both stars and want the mass ratio.
Key insight: In a binary, the more massive star moves more slowly around the center of mass.
Answer: The mass ratio follows the inverse speed ratio:
\[ \frac{M_2}{M_1} = \frac{v_1}{v_2} = \frac{25}{50} = 0.50. \]
So the second star has half the mass of the first, which means Star 1 is more massive.
Common misconception: The faster-moving star is not the more massive one. In binaries, the faster-moving star is the less massive one.
Problem 3: Mass-Luminosity Relation
Given: \(M = 4\,M_\odot\)
Find: \(L/L_\odot\)
Equation: \[ \frac{L}{L_\odot} = \left(\frac{M}{M_\odot}\right)^{3.5} \]
Solution:
\[ \frac{L}{L_\odot} = 4^{3.5} = 4^3 \times 4^{0.5} \]
\[ = 64 \times 2 = 128 \]
Unit check: This is a ratio of luminosities, so the result is dimensionless.
Sanity check: A \(4\,M_\odot\) main-sequence star should be much brighter than the Sun, so a factor of more than 100 is reasonable.
Answer: The star is about \(128\) times more luminous than the Sun.
Problem 4: Main-Sequence Lifetime
Given: \(M = 2.5\,M_\odot\)
Find: \(t_{\rm MS}\)
Equation: \[ \frac{t_{\rm MS}}{10\ \text{Gyr}} \approx \left(\frac{M}{M_\odot}\right)^{-2.5} \]
Solution:
\[ t_{\rm MS} \approx 10 \times (2.5)^{-2.5}\ \text{Gyr} \]
Since
\[ 2.5^{2.5} \approx 9.9, \]
we get
\[ t_{\rm MS} \approx \frac{10}{9.9}\ \text{Gyr} \approx 1.0\ \text{Gyr} \]
Unit check: The mass ratio is dimensionless, so the final unit stays in gigayears.
Sanity check: A star more massive than the Sun should live a shorter time, so about \(1\) Gyr is consistent with stellar-evolution trends.
Answer: The main-sequence lifetime is about \(1\ \text{Gyr}\), or 1 billion years.
Problem 5: Infer Mass from Luminosity
Given: \(L = 200\,L_\odot\)
Find: \(M/M_\odot\)
Equation: \[ \frac{L}{L_\odot} = \left(\frac{M}{M_\odot}\right)^{3.5} \]
So
\[ \frac{M}{M_\odot} = \left(\frac{L}{L_\odot}\right)^{1/3.5} \]
Solution:
\[ \frac{M}{M_\odot} = 200^{1/3.5} \approx 4.5 \]
Unit check: Because we are solving for a mass ratio, the result is dimensionless.
Sanity check: A star that is 200 times as luminous as the Sun should be several times more massive than the Sun, not just slightly more massive.
Answer: The star’s mass is about \(4.5\,M_\odot\).
Problem 6: Eclipse Depths in an Eclipsing Binary
Restatement: One eclipse is deep and the other is shallow. We want to interpret what that says about the stars and identify what extra data would let us measure precise masses.
Key insight: Different eclipse depths mean the two stars do not contribute equal amounts of light. The light curve also tells us the system is nearly edge-on, but masses still need orbital-speed information.
Answer: Unequal eclipse depths tell us the stars have different surface brightnesses, which usually means different temperatures and luminosities. The deeper eclipse happens when the brighter star is blocked, because the total light drops more.
To get precise masses, we also want radial velocity measurements. Those give the orbital speeds, and when combined with the nearly edge-on geometry from eclipses, they let us determine the stellar masses much more accurately.
Common misconception: An eclipsing light curve by itself is not enough to get precise masses. It gives geometry, but not the full dynamical information.
Challenge Problems
Challenge 1: Individual Masses in a Spectroscopic Binary
Given: - \(P = 5\) days \(= \dfrac{5}{365.25} \approx 0.0137\ \text{yr}\) - \(a = 0.074\ \text{AU}\) - \(v_1 = 100\ \text{km/s}\) - \(v_2 = 60\ \text{km/s}\)
Find: The individual masses \(M_1\) and \(M_2\)
Equation: \[ M_{\rm tot} = \frac{a^3}{P^2}, \qquad \frac{M_2}{M_1} = \frac{v_1}{v_2} \]
Solution:
First compute the total mass:
\[ M_{\rm tot} = \frac{(0.074)^3}{(0.0137)^2} \]
\[ M_{\rm tot} \approx \frac{4.05 \times 10^{-4}}{1.88 \times 10^{-4}} \approx 2.16\,M_\odot \]
Now use the velocity ratio:
\[ \frac{M_2}{M_1} = \frac{100}{60} = \frac{5}{3} \]
So
\[ M_2 = \frac{5}{3}M_1 \]
and
\[ M_1 + M_2 = M_1 + \frac{5}{3}M_1 = \frac{8}{3}M_1 = 2.16\,M_\odot \]
Solve for \(M_1\):
\[ M_1 = 2.16 \times \frac{3}{8} \approx 0.81\,M_\odot \]
Then
\[ M_2 = 2.16 - 0.81 \approx 1.35\,M_\odot \]
Unit check: The solar-unit form of Kepler’s law returns the total mass in \(M_\odot\), and the velocity ratio is dimensionless.
Sanity check: The star moving faster should be less massive. Here the star with \(100\ \text{km/s}\) ends up less massive than the star with \(60\ \text{km/s}\), which is exactly what we expect.
Answer: The individual masses are about \(0.81\,M_\odot\) and \(1.35\,M_\odot\).
Challenge 2: Two Equal-Mass Stars
Given: Each star has \(M = 2\,M_\odot\)
Find: The combined luminosity relative to the Sun
Equation: \[ \frac{L}{L_\odot} = \left(\frac{M}{M_\odot}\right)^{3.5} \]
Solution:
Each star has luminosity
\[ \frac{L}{L_\odot} = 2^{3.5} = 2^3 \times 2^{0.5} \approx 8 \times 1.41 \approx 11.3 \]
For two identical stars,
\[ L_{\rm total} \approx 2 \times 11.3\,L_\odot \approx 22.6\,L_\odot \]
Unit check: Luminosity ratios are dimensionless, so the total remains in units of \(L_\odot\).
Sanity check: Two equal stars should produce exactly twice the luminosity of one star, so doubling \(11.3\) to get about \(22.6\) is consistent.
Answer: Together the two stars are about 23 times as luminous as the Sun.
Challenge 3: Why Sirius B Does Not Follow the Main-Sequence Relation
Restatement: We want to explain why the mass-luminosity relation cannot be applied to Sirius B.
Key insight: The mass-luminosity relation is an empirical rule for main-sequence stars, whose luminosities are set by steady hydrogen fusion in their cores.
Answer: Sirius B is a white dwarf, not a main-sequence star. It is no longer powered by core hydrogen fusion. Its luminosity comes from leftover thermal energy, not from the same fusion-controlled physics that determines the luminosities of main-sequence stars.
So we should not expect Sirius B to follow the main-sequence mass-luminosity relation.
Common misconception: Students sometimes treat the mass-luminosity relation as universal. It is not. It applies only to main-sequence stars.
Challenge 4: Why Binary-Star Masses Calibrate Single Stars
Restatement: We want to connect binary-star measurements to the mass estimates astronomers make for single stars.
Key insight: Binary stars are the systems where stellar mass can be measured dynamically rather than inferred indirectly.
Answer: In binary systems, astronomers can combine orbital period, separation, and radial velocities to determine stellar masses using Newton’s version of Kepler’s third law. Those are the most direct mass measurements we can usually get for stars.
Once astronomers have reliable masses for many main-sequence stars in binaries, they can compare those masses with the stars’ luminosities and build the empirical mass-luminosity relation. After that calibration is established, the luminosities of single main-sequence stars can be used to estimate their masses.
So binary stars are foundational because they provide the calibration sample that makes the mass-luminosity relation possible.
In Observable \(\to\) Model \(\to\) Inference form: - Observable: binary periods, separations, and radial velocities - Model: Newtonian gravity applied to orbital motion - Inference: stellar masses, which then calibrate the mass-luminosity relation for single stars
Common misconception: Astronomers do not start by assuming the mass-luminosity relation. They first calibrate it using stars whose masses can be measured directly in binaries.